Pertanyaan

x → 0 lim tan 2 x sin 2 x cos 2 x − 1 = ...

$x \to 0 lim \frac{cos 2 x - 1}{tan 2 x sin 2 x} = ...$

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Jawaban terverifikasi

Jawaban

jawaban yang benar adalah B.

Pembahasan

Ingatlah identitascosinus sudut rangkap serta sifat-sifat limit fungsi Dengan menggunakan sifat-sifat limit fungsi dan identitas tersebut, didapatkan : Maka, . Oleh karena itu, jawaban yang benar adalah B.

Ingatlah identitas cosinus sudut rangkap

cos space 2 x equals 1 minus 2 space sin squared space x

serta sifat-sifat limit fungsi

$limit as x rightwards arrow 0 of fraction numerator sin space a x over denominator tan space b x end fraction equals a over b$
$limit as x rightwards arrow 0 of fraction numerator sin space a x over denominator sin space b x end fraction equals a over b$

Dengan menggunakan sifat-sifat limit fungsi dan identitas tersebut, didapatkan :

$table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 0 of space fraction numerator cos space 2 x minus 1 over denominator tan space 2 x space sin space 2 x end fraction end cell equals cell limit as x rightwards arrow 0 of space fraction numerator open parentheses 1 minus 2 sin to the power of 2 space end exponent x close parentheses minus 1 over denominator tan space 2 x space sin space 2 x end fraction end cell row blank equals cell limit as x rightwards arrow 0 of space fraction numerator negative 2 sin to the power of 2 space end exponent x over denominator tan space 2 x space sin space 2 x end fraction end cell row blank equals cell negative 2 times limit as x rightwards arrow 0 of open parentheses fraction numerator sin space x over denominator tan space 2 x end fraction times fraction numerator sin space x over denominator sin space 2 x end fraction close parentheses end cell row blank equals cell negative 2 times limit as x rightwards arrow 0 of fraction numerator sin space x over denominator tan space 2 x end fraction times limit as x rightwards arrow 0 of fraction numerator sin space x over denominator sin space 2 x end fraction end cell row blank equals cell negative 2 times 1 half times 1 half end cell row blank equals cell fraction numerator negative 2 over denominator 4 end fraction end cell row blank equals cell negative 0 comma 5 end cell end table$

Maka, $limit as x rightwards arrow 0 of space fraction numerator cos space 2 x minus 1 over denominator tan space 2 x space sin space 2 x end fraction equals negative 0 comma 5$ .

Oleh karena itu, jawaban yang benar adalah B.